Market etiquette or good manners in a minefield - page 83
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The optimal cagey partitioning of a tick BP with threshold H, should be considered the partitioning where there is the minimum number of consecutive single-coloured shoulders of a series of transactions.
If the distribution of the resulting row of transactions is such that more than 50% of consecutive shoulders have different colours, then why NS at all?
That's right. Indeed, the case you have voiced corresponds to an inefficient market in which one can and should make money! If you plot the transaction series (RT) in isolation from terminal time (just the countdown of the series), the effect is most obvious:
Exactly such areas (and they look the same for H+/- strategies) are fished by TC based on Kagi-buildings described in Pastukhov's dissertation. But there is a problem related to low profitability (in comparison to brokerage commissions) of such TS. It's connected with the fact that the classic strategy uses the simplest and most accessible property of Kagi formation - the reversal of PT, but there are other regularities... that's what the NS should be able to identify!
there are other patterns... that's what the NS should be able to detect!
Can we go into more detail on this point? I can't think of anything else except shoulder length (if you think of NS and kagi)
So I'm kind of stumped:
Suppose there is an NS that is trained with a number of: +1,-1,+1,-1,+1,-1,-1.... (i.e. binary inputs). In this case I can guess with about 80% accuracy from three times what it will learn. The non-linearity of the NS is irrelevant.
I would like as much as you to know the answers to some questions!
Let's see how a binary NS works. Suppose we have a training vector. All that such a NS can do with it to minimize the output error is to calculate the probabilities of outcomes for all possible input combinations. For clarity, let us have three inputs, then all input combinations are reduced to the following patterns (for beauty we will go from +/-1 to 0/1):
000
001
010
011
100
101
110
111
Let the training vector P be several times longer than the number of inputs d, then the NS will simply calculate the probability p of dropping 1 on each pattern (the probability for zero is 1-p). But we can do this without NS! There is one subtlety here. Namely, what will you do when you don't encounter any combinations in the training vector? What will you assign in reality to this pattern? - Nothing! You will have to increase the length of vector P until you encounter it (the pattern). And it's not a fact that you have enough available data, or even if you do, it's not a fact that you won't manage to get out of the optimal learning length. You know what I mean? This is where the advantage of NS comes into play. It turns out that it doesn't need all the training (for all occasions), but is capable of generalizing available knowledge with maximum generalization reliability! In other words, it reconstructs the most likely outcome for a pattern by itself, even if it wasn't there before in the training program. It's like an adult - we don't need a precedent to make a decision in this or that situation.
So, it will make a decision by projecting the available data (input) onto a certain hyper-surface that it builds in the feature space during its training. This surface is multidimensional (by number of inputs) and can be a plane, or a surface of higher order (paraboloid, hyperboloid for three dimensions, etc.). The presence of non-linearity, allows for complex surface topologies, and it doesn't matter that the input is binary, what matters is that it is projected onto a non-trivial surface.
So, I think non-linearity even with a binary input plays a role.
I'm a little confused about the error calculation for the hidden layer with non-linear FA. Can you please check if I am taking the hidden layer error correctly?
Here I am confused by the fact that the error at the output of the hidden layer is equal to the value of the microcorrection of the corresponding output layer synapse
No, it's not all right!
1. Find the NS output - OUT. You've got that right.
2. Calculate the error of the NS: dOUT=x-OUT. Knowing this error, find the value of correction weights of the output neuron. This is also correct.
3. the same error is at the output of each i-th neuron of the hidden layer (input layer), recalculate it to the input by the formula: dIn[i]=dOUT*(1-out[i]^2), where out[i] is output of the i-th neuron of the hidden layer. Knowing the error brought to the input of each neuron(dIn[i]), you find the value of correction weights in the input layer.
I was giving 100 - 120 epochs for a single layer. For a two-layer, is it probably not enough? It doesn't give good results on the kotier(clockwork) yet.
Something about her on the kotier(clockwork) is not giving good results so far.
Expecting something different?
Did you expect something different?
Honestly, yeah. I mean, the single-layer thing was working. You're right, though, you should stop with the timeframes for good.
I was giving 100 - 120 epochs for a single layer. For a two-layer, is it probably not enough? It's not giving good results on the kotier (clockwork) yet.
I thought you said that about the single layer...
I've got a double layer on the chapel consistently giving th<=0.05, and a single layer around zero.